Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{-8z^2 - 48z}{z^2 - 5z - 36} \div \dfrac{z + 6}{z + 4} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-8z^2 - 48z}{z^2 - 5z - 36} \times \dfrac{z + 4}{z + 6} $ First factor the quadratic. $t = \dfrac{-8z^2 - 48z}{(z + 4)(z - 9)} \times \dfrac{z + 4}{z + 6} $ Then factor out any other terms. $t = \dfrac{-8z(z + 6)}{(z + 4)(z - 9)} \times \dfrac{z + 4}{z + 6} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -8z(z + 6) \times (z + 4) } { (z + 4)(z - 9) \times (z + 6) } $ $t = \dfrac{ -8z(z + 6)(z + 4)}{ (z + 4)(z - 9)(z + 6)} $ Notice that $(z + 6)$ and $(z + 4)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ -8z(z + 6)\cancel{(z + 4)}}{ \cancel{(z + 4)}(z - 9)(z + 6)} $ We are dividing by $z + 4$ , so $z + 4 \neq 0$ Therefore, $z \neq -4$ $t = \dfrac{ -8z\cancel{(z + 6)}\cancel{(z + 4)}}{ \cancel{(z + 4)}(z - 9)\cancel{(z + 6)}} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $t = \dfrac{-8z}{z - 9} ; \space z \neq -4 ; \space z \neq -6 $